Churchill Friction Factor Equation question
 

In "A" sub equation - A=(2.4571n(1/(7/Re)^0.9+(0.27e/D)))^16 - what does the "n" stand for?

Thanks,

Nick

Re: Churchill Friction Factor Equation question
 

Is there any chance is says: A = (2.457 ln(1/(7/Re)^0.9 + (0.27e/D)))^16

and ln(...) the natural or Neperian logarithm

ln(...) = log(...)/log(exp(1))

Re: Churchill Friction Factor Equation question
 

Yep! That's exactly what it says. I just got an email back from some one, telling me that its not 1n, but its ln. In the font in which I read the equation, the 1 and the l were all but identical.

Thanks!

Nick